How to find maximum and minimum values in complex analysis??

Question

Another text book question that I have the answer for but not sure what formula to use?

Consider the function f(z)= (z + 1)^2 and the closed triangular region R with verticies at the points z=0,2,i

Find points in R where |f(z)| has its maximum and minimum values

I know two things. I have to interpret |f(z)| as the square of the distance between z and -1

And I know there are two values for the answer z=2 and z=0

How on earth do I get a real numbered answer for this quesiton?? Can you please help me understand the concept?

Thanks guys! :)

Answer 1

Let h(z) = |f(z). Then, h is a real valued function defined in the region R of the complex plane. If z = x + y i, x and y real, then h(z) = |(z + 1)^2| = |z + 1| ^2. Since this is a quadratic function. minimizing or mazimizing |z + 1| ^2 is the same as minimizing or mazimizing g(z) = |z +1| = |(x +1) + yi)| = (x+1)^2 + y^2

But since z is in the region R, there are the constraints

0 <= x <=2
y <= 1 - x/2,


| i
| R
|_________
0 2


So, you have 2 optmization problems:

maximize (minimize)

(x+1)^2 + y^2

subject to

0 <= x <=2
y <= 1 - x/2, x and y real

It's easy to see the minimum occurs when x=0 and y =0, that is z =0. This a point of R,

For a fixed x, we see the maximum occurs when y reaches its maximum feasible value, that is, y = 1 - x/2 So, we have to mazximize g(z) = (x +1)^2 + (1 - x/2)^2, a qudratic expression whose derivative with respect to x is g'(z) = 2(x +1) + 2(1 - x/2)* (-1/2) = 2x + 2 -1 +x/2 = 3x/2 -1. So, g has a minimum at x =2/3 ,which implies its maximum occurs at x = 0 0r at x =2. So, a simple calculation shows that maximum is at x =2 and y =0, that is, z =2.

Answer 2

Well, I've forgotten most of my complex analysis.

But just look at it as a function in the x,y plane. For convenience, you could translate everything so you're just looking at the distance from the origin, but you don't have to.

Wave your hands, and it's obvious that the maxima and minima will occur on the boundary of the region. That consists of three line segments. On each of those it's easy to reduce a function in the two variables x and y down to one variable.

And from there you can just proceed as you would in first-year calculus.